Question

How many milliliters of a 70.0 ml solution of 1.51 m bacl2 must be used to make 12.0 ml of a solution that has a concentration of 0.300 m cl–?

Answer 1

`1.43 \; \text{ml}`

Step-by-step explanation

• Number of moles of chloride ions in the final solution:
  `n = c \cdot V = 0.012 * 0.300 = 0.0036 \; \text{mol}`

• Each mole of barium chloride contains two moles of chloride ions. Thus the concentration of chloride ions in the initial 1.51 M barium chloride solution:
  `\begin{array}{lll} c(\text{Cl}^(-)) &=& 2 * c(\text{BaCl}_2)\\& =& 2.52 \; \text{mol} \cdot \text{L}^(-1) \\ & = &0.00252 \; \text{mol}\cdot \text{mL}^(-1) \end{array}`

Therefore

`\begin{array}{lll} V & = & n /c \\ & = & 0.0036 / 0.00252\\ & = & 1.43 \; \text{mL}\end{array}`